Question

In Young's double-slit experiment, the separation between the slits is $d$, distance between the slit and screen is $D(D>>d)$. In the interference pattern, there is a maxima exactly in front of each slit. Then the possible wavelength(s) used in the experiment are

• A. $d^2/D,d^2/2D,d^2/3D$
• B. $d^2/D,d^2/3D,d^2/5D$
• C. $d^2/2D,d^2/4D,d^2/6D$
• D. none of these

Answer 1

The correct option is C $d^2/2D,d^2/4D,d^2/6D$

path difference is $\sqrt{(}{D}^2+d^2)-D=n\lambda$ (as bright fringe is formed )
so ,
$D(1+\frac{d}{D}{)}^{\frac{1}{2}}-D=n\lambda$
$D+\frac{d^2}{2D}-D=n\lambda$
$\lambda =\frac{d^2}{2nD}$
the value of $n$ can be 1,2,3...
option $C$ is correct